Consider the right triangle with lengths 20, 21, 29 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{20}{29}\)
\(\cos(\theta) = \frac{21}{29}\)
\(\tan(\theta) = \frac{20}{21}\)
\(\csc(\theta) = \frac{29}{20}\)
\(\sec(\theta) = \frac{29}{21}\)
\(\cot(\theta) = \frac{21}{20}\)
\(\sin(\phi) = \frac{21}{29}\)
\(\cos(\phi) = \frac{20}{29}\)
\(\tan(\phi) = \frac{21}{20}\)
\(\csc(\phi) = \frac{29}{21}\)
\(\sec(\phi) = \frac{29}{20}\)
\(\cot(\phi) = \frac{20}{21}\)
Question
Consider the right triangle with lengths 5, 12, 13 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{5}{13}\)
\(\cos(\theta) = \frac{12}{13}\)
\(\tan(\theta) = \frac{5}{12}\)
\(\csc(\theta) = \frac{13}{5}\)
\(\sec(\theta) = \frac{13}{12}\)
\(\cot(\theta) = \frac{12}{5}\)
\(\sin(\phi) = \frac{12}{13}\)
\(\cos(\phi) = \frac{5}{13}\)
\(\tan(\phi) = \frac{12}{5}\)
\(\csc(\phi) = \frac{13}{12}\)
\(\sec(\phi) = \frac{13}{5}\)
\(\cot(\phi) = \frac{5}{12}\)
Question
Consider the right triangle with lengths 12, 35, 37 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{12}{37}\)
\(\cos(\theta) = \frac{35}{37}\)
\(\tan(\theta) = \frac{12}{35}\)
\(\csc(\theta) = \frac{37}{12}\)
\(\sec(\theta) = \frac{37}{35}\)
\(\cot(\theta) = \frac{35}{12}\)
\(\sin(\phi) = \frac{35}{37}\)
\(\cos(\phi) = \frac{12}{37}\)
\(\tan(\phi) = \frac{35}{12}\)
\(\csc(\phi) = \frac{37}{35}\)
\(\sec(\phi) = \frac{37}{12}\)
\(\cot(\phi) = \frac{12}{35}\)
Question
Consider the right triangle with lengths 11, 60, 61 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{11}{61}\)
\(\cos(\theta) = \frac{60}{61}\)
\(\tan(\theta) = \frac{11}{60}\)
\(\csc(\theta) = \frac{61}{11}\)
\(\sec(\theta) = \frac{61}{60}\)
\(\cot(\theta) = \frac{60}{11}\)
\(\sin(\phi) = \frac{60}{61}\)
\(\cos(\phi) = \frac{11}{61}\)
\(\tan(\phi) = \frac{60}{11}\)
\(\csc(\phi) = \frac{61}{60}\)
\(\sec(\phi) = \frac{61}{11}\)
\(\cot(\phi) = \frac{11}{60}\)
Question
Consider the right triangle with lengths 28, 45, 53 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{28}{53}\)
\(\cos(\theta) = \frac{45}{53}\)
\(\tan(\theta) = \frac{28}{45}\)
\(\csc(\theta) = \frac{53}{28}\)
\(\sec(\theta) = \frac{53}{45}\)
\(\cot(\theta) = \frac{45}{28}\)
\(\sin(\phi) = \frac{45}{53}\)
\(\cos(\phi) = \frac{28}{53}\)
\(\tan(\phi) = \frac{45}{28}\)
\(\csc(\phi) = \frac{53}{45}\)
\(\sec(\phi) = \frac{53}{28}\)
\(\cot(\phi) = \frac{28}{45}\)
Question
Consider the right triangle with lengths 8, 15, 17 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{8}{17}\)
\(\cos(\theta) = \frac{15}{17}\)
\(\tan(\theta) = \frac{8}{15}\)
\(\csc(\theta) = \frac{17}{8}\)
\(\sec(\theta) = \frac{17}{15}\)
\(\cot(\theta) = \frac{15}{8}\)
\(\sin(\phi) = \frac{15}{17}\)
\(\cos(\phi) = \frac{8}{17}\)
\(\tan(\phi) = \frac{15}{8}\)
\(\csc(\phi) = \frac{17}{15}\)
\(\sec(\phi) = \frac{17}{8}\)
\(\cot(\phi) = \frac{8}{15}\)
Question
Consider the right triangle with lengths 65, 72, 97 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{65}{97}\)
\(\cos(\theta) = \frac{72}{97}\)
\(\tan(\theta) = \frac{65}{72}\)
\(\csc(\theta) = \frac{97}{65}\)
\(\sec(\theta) = \frac{97}{72}\)
\(\cot(\theta) = \frac{72}{65}\)
\(\sin(\phi) = \frac{72}{97}\)
\(\cos(\phi) = \frac{65}{97}\)
\(\tan(\phi) = \frac{72}{65}\)
\(\csc(\phi) = \frac{97}{72}\)
\(\sec(\phi) = \frac{97}{65}\)
\(\cot(\phi) = \frac{65}{72}\)
Question
Consider the right triangle with lengths 3, 4, 5 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{3}{5}\)
\(\cos(\theta) = \frac{4}{5}\)
\(\tan(\theta) = \frac{3}{4}\)
\(\csc(\theta) = \frac{5}{3}\)
\(\sec(\theta) = \frac{5}{4}\)
\(\cot(\theta) = \frac{4}{3}\)
\(\sin(\phi) = \frac{4}{5}\)
\(\cos(\phi) = \frac{3}{5}\)
\(\tan(\phi) = \frac{4}{3}\)
\(\csc(\phi) = \frac{5}{4}\)
\(\sec(\phi) = \frac{5}{3}\)
\(\cot(\phi) = \frac{3}{4}\)
Question
Consider the right triangle with lengths 11, 60, 61 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{11}{61}\)
\(\cos(\theta) = \frac{60}{61}\)
\(\tan(\theta) = \frac{11}{60}\)
\(\csc(\theta) = \frac{61}{11}\)
\(\sec(\theta) = \frac{61}{60}\)
\(\cot(\theta) = \frac{60}{11}\)
\(\sin(\phi) = \frac{60}{61}\)
\(\cos(\phi) = \frac{11}{61}\)
\(\tan(\phi) = \frac{60}{11}\)
\(\csc(\phi) = \frac{61}{60}\)
\(\sec(\phi) = \frac{61}{11}\)
\(\cot(\phi) = \frac{11}{60}\)
Question
Consider the right triangle with lengths 3, 4, 5 and acute angles \(\phi\) and \(\theta\), where \(\theta<\phi\). Find all the trigonometric ratios.
\(\sin(\theta)\) = /
\(\cos(\theta)\) = /
\(\tan(\theta)\) = /
\(\csc(\theta)\) = /
\(\sec(\theta)\) = /
\(\cot(\theta)\) = /
\(\sin(\phi)\) = /
\(\cos(\phi)\) = /
\(\tan(\phi)\) = /
\(\csc(\phi)\) = /
\(\sec(\phi)\) = /
\(\cot(\phi)\) = /
Solution
This is definitional. It is also exhaustive and tediuous. Sorry?
\(\sin(\theta) = \frac{3}{5}\)
\(\cos(\theta) = \frac{4}{5}\)
\(\tan(\theta) = \frac{3}{4}\)
\(\csc(\theta) = \frac{5}{3}\)
\(\sec(\theta) = \frac{5}{4}\)
\(\cot(\theta) = \frac{4}{3}\)
\(\sin(\phi) = \frac{4}{5}\)
\(\cos(\phi) = \frac{3}{5}\)
\(\tan(\phi) = \frac{4}{3}\)
\(\csc(\phi) = \frac{5}{4}\)
\(\sec(\phi) = \frac{5}{3}\)
\(\cot(\phi) = \frac{3}{4}\)
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.22 radians.
The adventurer walks toward the monument, getting 1200 meters closer; the viewing angle increases to 1.45 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.22) = \frac{h}{1200+z}\]
\[\tan(1.45) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.2236194 = \frac{h}{1200+z}\]
\[8.2380928 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 8.2380928. Remember, dividing by 8.2380928 is equivalent to multiplying by \(\frac{1}{8.2380928}\), which is equivalent to multiplying by \(0.1213873\).
\[z ~=~ \frac{h}{8.2380928}\]
\[z ~=~ \frac{1}{8.2380928} h\]
\[z ~=~ 0.1213873 h\]
Substitute \(0.1213873 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 275.8306069\]
Round to 2 significant digits.
\[h = 280\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.25 radians.
The adventurer walks toward the monument, getting 260 meters closer; the viewing angle increases to 0.87 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.25) = \frac{h}{260+z}\]
\[\tan(0.87) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.2553419 = \frac{h}{260+z}\]
\[1.1853249 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 1.1853249. Remember, dividing by 1.1853249 is equivalent to multiplying by \(\frac{1}{1.1853249}\), which is equivalent to multiplying by \(0.8436506\).
\[z ~=~ \frac{h}{1.1853249}\]
\[z ~=~ \frac{1}{1.1853249} h\]
\[z ~=~ 0.8436506 h\]
Substitute \(0.8436506 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 84.6170502\]
Round to 2 significant digits.
\[h = 85\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.17 radians.
The adventurer walks toward the monument, getting 520 meters closer; the viewing angle increases to 0.76 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.17) = \frac{h}{520+z}\]
\[\tan(0.76) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.1716568 = \frac{h}{520+z}\]
\[0.9504515 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 0.9504515. Remember, dividing by 0.9504515 is equivalent to multiplying by \(\frac{1}{0.9504515}\), which is equivalent to multiplying by \(1.0521316\).
\[z ~=~ \frac{h}{0.9504515}\]
\[z ~=~ \frac{1}{0.9504515} h\]
\[z ~=~ 1.0521316 h\]
Substitute \(1.0521316 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 108.9359942\]
Round to 2 significant digits.
\[h = 110\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.3 radians.
The adventurer walks toward the monument, getting 980 meters closer; the viewing angle increases to 0.76 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.3) = \frac{h}{980+z}\]
\[\tan(0.76) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.3093362 = \frac{h}{980+z}\]
\[0.9504515 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 0.9504515. Remember, dividing by 0.9504515 is equivalent to multiplying by \(\frac{1}{0.9504515}\), which is equivalent to multiplying by \(1.0521316\).
\[z ~=~ \frac{h}{0.9504515}\]
\[z ~=~ \frac{1}{0.9504515} h\]
\[z ~=~ 1.0521316 h\]
Substitute \(1.0521316 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 449.4183003\]
Round to 2 significant digits.
\[h = 450\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.17 radians.
The adventurer walks toward the monument, getting 260 meters closer; the viewing angle increases to 0.49 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.17) = \frac{h}{260+z}\]
\[\tan(0.49) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.1716568 = \frac{h}{260+z}\]
\[0.5333881 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 0.5333881. Remember, dividing by 0.5333881 is equivalent to multiplying by \(\frac{1}{0.5333881}\), which is equivalent to multiplying by \(1.8748073\).
\[z ~=~ \frac{h}{0.5333881}\]
\[z ~=~ \frac{1}{0.5333881} h\]
\[z ~=~ 1.8748073 h\]
Substitute \(1.8748073 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 65.8099647\]
Round to 2 significant digits.
\[h = 66\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.25 radians.
The adventurer walks toward the monument, getting 1700 meters closer; the viewing angle increases to 0.98 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.25) = \frac{h}{1700+z}\]
\[\tan(0.98) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.2553419 = \frac{h}{1700+z}\]
\[1.4909583 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 1.4909583. Remember, dividing by 1.4909583 is equivalent to multiplying by \(\frac{1}{1.4909583}\), which is equivalent to multiplying by \(0.6707096\).
\[z ~=~ \frac{h}{1.4909583}\]
\[z ~=~ \frac{1}{1.4909583} h\]
\[z ~=~ 0.6707096 h\]
Substitute \(0.6707096 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 523.7847933\]
Round to 2 significant digits.
\[h = 520\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.32 radians.
The adventurer walks toward the monument, getting 1300 meters closer; the viewing angle increases to 0.89 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.32) = \frac{h}{1300+z}\]
\[\tan(0.89) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.3313894 = \frac{h}{1300+z}\]
\[1.2345995 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 1.2345995. Remember, dividing by 1.2345995 is equivalent to multiplying by \(\frac{1}{1.2345995}\), which is equivalent to multiplying by \(0.8099793\).
\[z ~=~ \frac{h}{1.2345995}\]
\[z ~=~ \frac{1}{1.2345995} h\]
\[z ~=~ 0.8099793 h\]
Substitute \(0.8099793 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 588.869812\]
Round to 2 significant digits.
\[h = 590\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.28 radians.
The adventurer walks toward the monument, getting 880 meters closer; the viewing angle increases to 0.96 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.28) = \frac{h}{880+z}\]
\[\tan(0.96) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.2875543 = \frac{h}{880+z}\]
\[1.4283575 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 1.4283575. Remember, dividing by 1.4283575 is equivalent to multiplying by \(\frac{1}{1.4283575}\), which is equivalent to multiplying by \(0.7001048\).
\[z ~=~ \frac{h}{1.4283575}\]
\[z ~=~ \frac{1}{1.4283575} h\]
\[z ~=~ 0.7001048 h\]
Substitute \(0.7001048 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 316.8318087\]
Round to 2 significant digits.
\[h = 320\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.33 radians.
The adventurer walks toward the monument, getting 1200 meters closer; the viewing angle increases to 0.72 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.33) = \frac{h}{1200+z}\]
\[\tan(0.72) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.3425249 = \frac{h}{1200+z}\]
\[0.8770679 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 0.8770679. Remember, dividing by 0.8770679 is equivalent to multiplying by \(\frac{1}{0.8770679}\), which is equivalent to multiplying by \(1.1401626\).
\[z ~=~ \frac{h}{0.8770679}\]
\[z ~=~ \frac{1}{0.8770679} h\]
\[z ~=~ 1.1401626 h\]
Substitute \(1.1401626 h\) for \(z\) in the first equation of the system.
If you plug in the known values, you should be able to get \(h\).
\[h = 674.4098373\]
Round to 2 significant digits.
\[h = 670\]
Question
An adventurer, crossing a vast (flat) desert, approaches a distant monument. When the adventurer first notices the monument, the viewing angle (the angle of elevation to the top of the monument) is 0.13 radians.
The adventurer walks toward the monument, getting 1300 meters closer; the viewing angle increases to 1.43 radians.
Below are two methods. The first uses long decimal representations to somewhat simplify the algebra. The second method abstracts the problem further to derive a formula.
Long decimals
First, label some unknowns with variables.
Write a system of equations.
\[\tan(0.13) = \frac{h}{1300+z}\]
\[\tan(1.43) = \frac{h}{z}\]
Evaluate decimal approximations.
\[0.1307373 = \frac{h}{1300+z}\]
\[7.0554638 = \frac{h}{z}\]
Isolate (solve for) \(z\) in the second equation by multiplying both sides by \(z\) and dividing both sides by 7.0554638. Remember, dividing by 7.0554638 is equivalent to multiplying by \(\frac{1}{7.0554638}\), which is equivalent to multiplying by \(0.1417341\).
\[z ~=~ \frac{h}{7.0554638}\]
\[z ~=~ \frac{1}{7.0554638} h\]
\[z ~=~ 0.1417341 h\]
Substitute \(0.1417341 h\) for \(z\) in the first equation of the system.
Solve those equations.
\[p = 9.023801\]\[q = 6.8902263\]
Add the parts.
\[x = 9.023801+6.8902263\]\[x = 15.9140274\]
Question
A ship sailed 7.4 miles in a direction of 30° North of East. The ship turned and sailed 9.4 miles in a direction 74° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 7.8 miles in a direction of 19° North of East. The ship turned and sailed 6.7 miles in a direction 41° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 8.5 miles in a direction of 13° North of East. The ship turned and sailed 6.4 miles in a direction 38° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 5.5 miles in a direction of 30° North of East. The ship turned and sailed 4.7 miles in a direction 71° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 9.3 miles in a direction of 28° North of East. The ship turned and sailed 9.8 miles in a direction 74° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 5.6 miles in a direction of 16° North of East. The ship turned and sailed 4.9 miles in a direction 63° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 7.2 miles in a direction of 16° North of East. The ship turned and sailed 6.6 miles in a direction 54° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 2.9 miles in a direction of 22° North of East. The ship turned and sailed 3.5 miles in a direction 73° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 9.3 miles in a direction of 30° North of East. The ship turned and sailed 7.6 miles in a direction 55° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
A ship sailed 2.7 miles in a direction of 27° North of East. The ship turned and sailed 3.3 miles in a direction 53° North of East.
The ship’s captain suffers a terrible accident. The first officer radios for help and sets anchor, waiting for help while administering first aid.
A helicopter, starting where the ship started, wishes to fly directly to the ship. What direction (\(\theta\) rounded to nearest degree) and distance (\(r\) rounded to nearest tenth of a mile) should the helicopter fly?
The direction: \(\theta =\)
The distance: \(r =\)
Solution
Start by naming some lengths (the legs of the two small triangles).
Evaluate \(a\), \(b\), \(j\), and \(k\). Remember to set your calculator to DEGREE MODE.
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 9.5% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 9.5 feet. We wish to determine the angle of inclination (in degrees) of a 9.5% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{9.5}{100}\right)\]
\[\theta \approx 5.4^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 21.3% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 21.3 feet. We wish to determine the angle of inclination (in degrees) of a 21.3% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{21.3}{100}\right)\]
\[\theta \approx 12^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 10.7% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 10.7 feet. We wish to determine the angle of inclination (in degrees) of a 10.7% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{10.7}{100}\right)\]
\[\theta \approx 6.1^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 26.5% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 26.5 feet. We wish to determine the angle of inclination (in degrees) of a 26.5% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{26.5}{100}\right)\]
\[\theta \approx 14.8^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 28.6% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 28.6 feet. We wish to determine the angle of inclination (in degrees) of a 28.6% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{28.6}{100}\right)\]
\[\theta \approx 16^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 28.9% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 28.9 feet. We wish to determine the angle of inclination (in degrees) of a 28.9% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{28.9}{100}\right)\]
\[\theta \approx 16.1^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 19.1% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 19.1 feet. We wish to determine the angle of inclination (in degrees) of a 19.1% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{19.1}{100}\right)\]
\[\theta \approx 10.8^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 7.1% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 7.1 feet. We wish to determine the angle of inclination (in degrees) of a 7.1% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{7.1}{100}\right)\]
\[\theta \approx 4.1^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 23.8% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 23.8 feet. We wish to determine the angle of inclination (in degrees) of a 23.8% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{23.8}{100}\right)\]
\[\theta \approx 13.4^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
On steep streets, you may see signs that indicate the grade, which is usually shown as a percentage. The grade is equivalent to slope, written as a percentage.
The highest grade of the Ashuwillticook Rail Trail is 2.5%. The Mohawk Trail (Route 2), near the hairpin turn, has sections of 8% grade. Notch Road, heading up Mount Greylock, has grades as high as 12%. Circuit Road, below Berry Pond in Pittsfield State Forest, has a section of 16% grade. Thunderbolt Ski Trail’s steepest section has 35% grade.
A grade of 39.8% implies that for every 100 feet of horizontal travel (as seen on a map), the elevation changes by 39.8 feet. We wish to determine the angle of inclination (in degrees) of a 39.8% grade.
Find \(\theta\), the angle of inclination in degrees. The tolerance is \(\pm0.1^\circ\).
Solution
Remember to use DEGREE mode.
\[\theta = \arctan\left(\frac{39.8}{100}\right)\]
\[\theta \approx 21.7^\circ\]
Humans tend to think terrain is much steeper than it truly is. It is tempting to think a hill is going up at a 45\(^\circ\) angle, but this would be a 100% grade, much steeper than any public street in the world. The steepest public roads have angles of inclination under \(22^\circ\) (under 40% grade). Some examples: Canton Avenue in Pennsylvania, Baldwin Street in New Zealand, and Ffordd_Pen_Llech in North Wales.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 3.53\]\[\theta=\frac{15\pi}{13}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{15\pi}{13}-\pi\right|\]
\[\phi ~=~ \frac{2\pi}{13}\]
plot of chunk unnamed-chunk-4
Remember, \(r=3.53\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 3, we know the signs of \(x\) and \(y\).
\[x = -3.1256598\]
\[y = -1.6404728\]
If you round to the hundredths place:
\[x = -3.13\]\[y = -1.64\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 3.17\]\[\theta=\frac{6\pi}{7}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{6\pi}{7}-\pi\right|\]
\[\phi ~=~ \frac{\pi}{7}\]
plot of chunk unnamed-chunk-4
Remember, \(r=3.17\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 2, we know the signs of \(x\) and \(y\).
\[x = -2.8560713\]
\[y = 1.3754115\]
If you round to the hundredths place:
\[x = -2.86\]\[y = 1.38\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 3.41\]\[\theta=\frac{7\pi}{9}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{7\pi}{9}-\pi\right|\]
\[\phi ~=~ \frac{2\pi}{9}\]
plot of chunk unnamed-chunk-4
Remember, \(r=3.41\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 2, we know the signs of \(x\) and \(y\).
\[x = -2.6122116\]
\[y = 2.1919057\]
If you round to the hundredths place:
\[x = -2.61\]\[y = 2.19\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 4.94\]\[\theta=\frac{12\pi}{7}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{12\pi}{7}-2\pi\right|\]
\[\phi ~=~ \frac{2\pi}{7}\]
plot of chunk unnamed-chunk-4
Remember, \(r=4.94\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 4, we know the signs of \(x\) and \(y\).
\[x = 3.0800396\]
\[y = -3.8622475\]
If you round to the hundredths place:
\[x = 3.08\]\[y = -3.86\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 3.14\]\[\theta=\frac{11\pi}{9}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{11\pi}{9}-\pi\right|\]
\[\phi ~=~ \frac{2\pi}{9}\]
plot of chunk unnamed-chunk-4
Remember, \(r=3.14\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 3, we know the signs of \(x\) and \(y\).
\[x = -2.4053796\]
\[y = -2.0183531\]
If you round to the hundredths place:
\[x = -2.41\]\[y = -2.02\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 3.33\]\[\theta=\frac{8\pi}{11}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{8\pi}{11}-\pi\right|\]
\[\phi ~=~ \frac{3\pi}{11}\]
plot of chunk unnamed-chunk-4
Remember, \(r=3.33\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 2, we know the signs of \(x\) and \(y\).
\[x = -2.1806862\]
\[y = 2.5166461\]
If you round to the hundredths place:
\[x = -2.18\]\[y = 2.52\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 5.66\]\[\theta=\frac{14\pi}{11}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{14\pi}{11}-\pi\right|\]
\[\phi ~=~ \frac{3\pi}{11}\]
plot of chunk unnamed-chunk-4
Remember, \(r=5.66\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 3, we know the signs of \(x\) and \(y\).
\[x = -3.7065118\]
\[y = -4.2775426\]
If you round to the hundredths place:
\[x = -3.71\]\[y = -4.28\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 4.54\]\[\theta=\frac{15\pi}{11}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{15\pi}{11}-\pi\right|\]
\[\phi ~=~ \frac{4\pi}{11}\]
plot of chunk unnamed-chunk-4
Remember, \(r=4.54\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 3, we know the signs of \(x\) and \(y\).
\[x = -1.8859842\]
\[y = -4.1297293\]
If you round to the hundredths place:
\[x = -1.89\]\[y = -4.13\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 3.52\]\[\theta=\frac{14\pi}{11}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{14\pi}{11}-\pi\right|\]
\[\phi ~=~ \frac{3\pi}{11}\]
plot of chunk unnamed-chunk-4
Remember, \(r=3.52\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 3, we know the signs of \(x\) and \(y\).
\[x = -2.3051098\]
\[y = -2.6602385\]
If you round to the hundredths place:
\[x = -2.31\]\[y = -2.66\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 4.82\]\[\theta=\frac{9\pi}{8}\]
plot of chunk unnamed-chunk-2
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{9\pi}{8}-\pi\right|\]
\[\phi ~=~ \frac{\pi}{8}\]
plot of chunk unnamed-chunk-4
Remember, \(r=4.82\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 3, we know the signs of \(x\) and \(y\).
\[x = -4.4530993\]
\[y = -1.8445341\]
If you round to the hundredths place:
\[x = -4.45\]\[y = -1.84\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.43\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-4.34}{3.76}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Because the point is in quadrant II, we find the difference of \(\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = \pi-\phi\]
\[\theta\approx2.83\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{1.38}{-4.24}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Finding the reference angle by using the arctan function with the quotient of the absolute values of the rectangular coordinates.
\[\phi ~=~ \arctan\left(\frac{2.1}{2.63}\right)\]
\[\phi ~=~ 0.6738129\]
plot of chunk unnamed-chunk-4
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.61\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-2.1}{2.63}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Because the point is in quadrant II, we find the difference of \(\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = \pi-\phi\]
\[\theta\approx1.89\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{4.12}{-1.34}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Because the point is in quadrant II, we find the difference of \(\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = \pi-\phi\]
\[\theta\approx1.88\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{4.35}{-1.41}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.39\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-3.11}{2.48}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Because the point is in quadrant III, we add \(\pi\) and \(\phi\) to get \(\theta\).
\[\theta = \pi+\phi\]
\[\theta\approx4.35\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-3.33}{-1.26}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.03\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-4.42}{1.44}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.03\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-4.46}{1.45}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.14\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-3.67}{1.68}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function, so arctangent has a restricted range, from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\).
Find the long-leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the long-leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the short-leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the hypotenuse length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of 45°-45°-90° triangles and 30°-60°-90° triangles by using Pythagorean Theorem.
45-45-90
A 45°-45°-90° triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a 45-45-90 triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
30-60-90
The ratios of a 30-60-90 triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a 30-60-90 triangle. In that 30-60-90 triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{6}\) has hypotenuse length of 8.8 meters.
Find the long-leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{6}\) has long-leg length of 9.66 meters.
Find the short-leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{4}\) has hypotenuse length of 9.69 meters.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{4}\) has leg length of 6.86 meters.
Find the hypotenuse length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{4}\) has hypotenuse length of 8.38 meters.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{4}\) has leg length of 4.46 meters.
Find the hypotenuse length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{6}\) has long-leg length of 2.52 meters.
Find the short-leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{4}\) has hypotenuse length of 7.95 meters.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{4}\) has hypotenuse length of 6.34 meters.
Find the leg length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
A right triangle with an acute angle of \(\frac{\pi}{6}\) has long-leg length of 3.84 meters.
Find the hypotenuse length in meters. The tolerance is \(\pm0.01\) meters.
Solution
You should be able to derive the side-length ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles and \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangles by using Pythagorean Theorem.
Ratios of \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangles
\(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle’s ratios are easy to derive. Because it is isosceles, both legs are congruent. We can use \(x\) to represent the length of a leg and \(c\) to represent the length of the hypotenuse.
If we use Pythagorean Theorem:
\[x^2+x^2 = c^2\]
Combine similar terms.
\[2x^2 = c^2\]
Take the square root of both sides.
\[\sqrt{2x^2} = \sqrt{c^2}\]
Simplify.
\[\frac{c}{x} = \sqrt{2}\]
Any of the triangles below indicate the ratios of a \(\frac{\pi}{4}\)-\(\frac{\pi}{4}\)-\(\frac{\pi}{2}\) triangle.
The 3-part ratios can be expressed in various ways.
\[1:1:\sqrt{2}\]
\[\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1\]
\[\frac{\sqrt{2}}{2} : \frac{\sqrt{2}}{2} : 1\]
The \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle ratios
The ratios of a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle are also straight-forward to prove. Begin by drawing an equilateral triangle. Cut the equilateral triangle in half. That half of the equilateral triangle will be a \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle. In that \(\frac{\pi}{6}\)-\(\frac{\pi}{3}\)-\(\frac{\pi}{2}\) triangle, the short leg is half as long as the hypotenuse.
Use Pythagorean Theorem:
\[y^2+b^2 = (2y)^2\]
An exponent can distribute over multiplication.
\[y^2+b^2 = 4y^2\]
Subtract \(y^2\) from both sides.
\[b^2 = 3y^2\]
Square root both sides.
\[b = y\sqrt{3}\]
The 3-part ratios can be expressed in various ways.
